Computing $ I_n=\int_0^1 \frac{x^n}{6+x-x^2} d x $

Question

After reading the reduction formula in the post, I am curious about the closed form of the integral $$ I_n=\int_0^1 \frac{x^n}{6+x-x^2} d x $$ I first resolve the integrand into two partial fractions as $$ I_n=\frac{1}{5} \left[\underbrace{\int_0^1 \frac{x^n}{3-x} d x}_{J_n} +\underbrace{\int_0^1 \frac{x^n}{2+x} d x}_{K_n} \right] $$ Then I tackle the two integrals in general, $$ \begin{aligned} \int_0^1 \frac{x^n}{a-x} d x & =\int_{a-1}^a \frac{(a-x)^n}{x} d x \\ \\ & =\int_{a-1}^a\left(\frac{a^n}{x}-\sum_{k=1}^n\left(\begin{array}{c} n \\ k \end{array}\right) a^{n-k} x^{k-1}\right) dx \\ & =a^n \ln \left(\frac{a}{a-1}\right)-\sum_{k=1}^n\left(\begin{array}{l} n \\ k \end{array}\right) \frac{a^{n}-a^{n-k}(a-1)^k}{k} \end{aligned} $$ Hence $$J_n =3^n \ln \left(\frac{3}{2}\right)-\sum_{k=1}^n\left(\begin{array}{l} n \\ k \end{array}\right) \frac{3^{n}-3^{n-k}2^k}{k} $$ and $$K_n= -\left[(-2)^n \ln \left(\frac{2}{3}\right)-\sum_{k=1}^n\left(\begin{array}{l} n \\ k \end{array}\right) \frac{(-2)^{n}-(-1)^n2^{n-k}3^k}{k} \right]$$ We can now conclude that the closed form of the integral is $$ \begin{aligned}I_n= &\ \frac{1}{5} \left[ 3^n\left[\ln \left(\frac{3}{2}\right)-\sum_{k=1}^n\left(\begin{array}{l} n \\ k \end{array}\right) \frac{1-\left(\frac{2}{3}\right)^k}{k}\right]+ (-2)^{n}\left[\ln \left(\frac{3}{2}\right)-\sum_{k=1}^n\left(\begin{array}{l} n \\ k \end{array}\right) \frac{1-\left(\frac{3}{2}\right)^k}{k}\right]\right]\\=& \frac{3^n+(-2)^n}{5} \ln \left(\frac{3}{2}\right) -\frac{1}{5}\left[\sum_{k=1}^n\left(\begin{array}{l} n \\ k \end{array}\right) \frac{1}{k}\left(3^n\left(1-\left(\frac{2}{3}\right)^k\right)+(-2)^n\left(1-\left(\frac{3}{2}\right)^k\right)\right]\right. \end{aligned} $$

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My question:

Is there any alternative method? Your comments and alternative methods are highly appreciated.

Answer 1

Here is a method to arrive at a simpler closed-form result. Note that

\begin{align} J_n(a)=&\int_0^1 \frac{x^n}{x+a}dx = \frac1n - a J_{n-1}(a)\\ = &\ (-a)^n\ln\frac{a+1}a+\sum_{k=0}^{n-1}\frac{(-a)^k}{n-k} \end{align} Then \begin{align} &\int_0^1 \frac{x^n}{6+x-x^2} d x\\ =&\ \frac{1}{5} \int_0^1 \frac{x^n}{2+x}-\frac{x^n}{x-3}\ d x = \frac15[J_n(2)-J_n(-3)] \\ =&\ \frac{3^n+(-2)^n}5\ln\frac32 +\frac15 \sum_{k=0}^{n-1}\frac{(-2)^k-3^k}{n-k} \end{align}

Answer 2

Define the function $I : [1,\infty) \to \mathbb{R}_{\ge 0}$ by $I(t) = \int_{0}^{1} \frac{x^t}{6+x-x^2} dx$. I am aware that you are searching for a closed form of this integral, I think the best we have currently is @userrandrand 's answer in the comments and your solution. I present a numerical approximation which is close to the true solution and gets better with larger $t$. First observe that $0 \le x(1-x) \le \frac{1}{4}$ for $x\in [0,1]$, therefore $$\frac{4}{25(t+1)} = \int_{0}^{1} \frac{x^t}{6+\frac{1}{4}} dx \le I(t) \le \frac{1}{6}\int_{0}^{1} x^t dx = \frac{1}{6(t+1)}.$$ We can do even better by realizing that the contribution in the integral is only from a neighborhood near $1$, as $x^t$ is near zero away from $1$. This implies that we can observe the leading behaviour of the integral by taylor expanding the rest of the integrand at $x=1$. Taylor expanding $1/(6+x-x^2)$ near $x = 1$, we have $$\frac{1}{6+x-x^2} = \frac{1}{6}+\frac{1}{36}(x-1)+\frac{7}{216}\left(x-1\right)^{2} + O(x-1)^3.$$ With this, the integral can be approximated as following: \begin{align*} I(t) &\approx \int_{0}^{1} \left(\frac{1}{6}+\frac{1}{36}(x-1)+\frac{7}{216}\left(x-1\right)^{2}\right)x^t dx. \\ &= \frac{1}{216}\left(\frac{37}{t+1}-\frac{8}{t+2}+\frac{7}{t+3}\right). \end{align*} For $t=1$, you already have two digits of accuracy, which rapidly improves with a larger value of $t$. You can take more terms in the Taylor expansion for better accuracy, but the approximation becomes more convoluted.

Answer 3

Thank @userrandrand for introducing the LerchPhi function $\Phi(z, s, a) $with an integral representation as below: $$ \Phi(z, s, a)=\frac{1}{\Gamma(s)} \int_0^1 \frac{[-\ln (t)]^{s-1} t^{a-1}}{1-z t} d t, $$ we have $$ I_n=\frac{1}{5}\left[\frac{1}{3} \int_0^1 \frac{x^n}{1-\frac{1}{3} x} d x+\frac{1}{2} \int_0^1 \frac{x^n}{1-\left(-\frac{1}{2}\right) x} d x\right]= \frac{1}{30}\left[2 \Phi\left(\frac{1}{3}, 1, n+1\right)+3 \Phi\left(-\frac{1}{2}, 1, n+1\right)\right] $$